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Transcript
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“main”
2007/2/16
page 274
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CHAPTER 4
Vector Spaces
and setting s = 0 and t = 1 yields the linear combination
2v1 + v4 = 0.
(4.5.3)
We can solve (4.5.2) for v3 in terms of v1 and v2 , and we can solve (4.5.3) for v4 in terms
of v1 . Hence, according to Theorem 4.5.2, we have
span{v1 , v2 , v3 , v4 } = span{v1 , v2 }.
By Proposition 4.5.7, v1 and v2 are linearly independent, so {v1 , v2 } is the linearly
independent set we are seeking. Geometrically, the subspace of R3 spanned by v1 and
v2 is a plane, and the vectors v3 and v4 lie in this plane.
Linear Dependence and Linear Independence in Rn
Let {v1 , v2 , . . . , vk } be a set of vectors in Rn , and let A denote the matrix that has
v1 , v2 , . . . , vk as column vectors. Thus,
A = [v1 , v2 , . . . , vk ].
(4.5.4)
Since each of the given vectors is in Rn , it follows that A has n rows and is therefore an
n × k matrix.
The linear combination c1 v1 + c2 v2 + · · · + ck vk = 0 can be written in matrix form
as (see Theorem 2.2.9)
Ac = 0,
(4.5.5)
where A is given in Equation (4.5.4) and c = [c1 c2 . . . ck ]T . Consequently, we can
state the following theorem and corollary:
Theorem 4.5.14
Let v1 , v2 , . . . , vk be vectors in Rn and A = [v1 , v2 , . . . , vk ]. Then {v1 , v2 , . . . , vk } is
linearly dependent if and only if the linear system Ac = 0 has a nontrivial solution.
Corollary 4.5.15
Let v1 , v2 , . . . , vk be vectors in Rn and A = [v1 , v2 , . . . , vk ].
1. If k > n, then {v1 , v2 , . . . , vk } is linearly dependent.
2. If k = n, then {v1 , v2 , . . . , vk } is linearly dependent if and only if det(A) = 0.
Proof If k > n, the system (4.5.5) has an infinite number of solutions (see Corollary 2.5.11), hence the vectors are linearly dependent by Theorem 4.5.14.
On the other hand, if k = n, the system (4.5.5) is n × n, and hence, from Corollary
3.2.5, it has an infinite number of solutions if and only if det(A) = 0.
Example 4.5.16
Determine whether the given vectors are linearly dependent or linearly independent in
R4 .
1. v1 = (1, 3, −1, 0), v2 = (2, 9, −1, 3), v3 = (4, 5, 6, 11), v4 = (1, −1, 2, 5),
v5 = (3, −2, 6, 7).
2. v1 = (1, 4, 1, 7), v2 = (3, −5, 2, 3), v3 = (2, −1, 6, 9), v4 = (−2, 3, 1, 6).
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